The average translational kinetic energy of `O_(2)` (molar mass 32) molecules at a particular temperature is `0.048 eV`. The translational kinetic energy of `N_(2)` (molar mass 28) molecules in (eV) at the same temperature is (JEE 1997)
(a) 0.0015 (b) 0.003 ( c) 0.048 (d) 0.768

Explanation

To solve the problem, we need to understand the relationship between the average translational kinetic energy of gas molecules and their temperature. The average translational kinetic energy (KE) of a gas molecule is given by the formula:

\[ KE = \frac{3}{2} k T \]

where:
- \( k \) is the Boltzmann constant,
- \( T \) is the absolute temperature in Kelvin.

Given that the average translational kinetic energy of \( O_2 \) molecules at a certain temperature is \( 0.048 \, \text{eV} \), we can conclude that the average translational kinetic energy of any gas at that same temperature will also be the same, regardless of the molar mass of the gas.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Average translational kinetic energy of \( O_2 \) = \( 0.048 \, \text{eV} \)
- Molar mass of \( O_2 \) = 32
- Molar mass of \( N_2 \) = 28

2. **Understand the Concept**:
- The average translational kinetic energy of gas molecules at a given temperature is independent of the type of gas. It only depends on the temperature.

3. **Apply the Concept**:
- Since both gases are at the same temperature, the average translational kinetic energy of \( N_2 \) will also be equal to that of \( O_2 \).

4. **Conclusion**:
- Therefore, the average translational kinetic energy of \( N_2 \) at the same temperature is also \( 0.048 \, \text{eV} \).

### Final Answer:
The translational kinetic energy of \( N_2 \) molecules at the same temperature is \( 0.048 \, \text{eV} \).

### Correct Option:
(c) 0.048

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